Integrand size = 11, antiderivative size = 132 \[ \int x^7 (a+b x)^{10} \, dx=-\frac {a^7 (a+b x)^{11}}{11 b^8}+\frac {7 a^6 (a+b x)^{12}}{12 b^8}-\frac {21 a^5 (a+b x)^{13}}{13 b^8}+\frac {5 a^4 (a+b x)^{14}}{2 b^8}-\frac {7 a^3 (a+b x)^{15}}{3 b^8}+\frac {21 a^2 (a+b x)^{16}}{16 b^8}-\frac {7 a (a+b x)^{17}}{17 b^8}+\frac {(a+b x)^{18}}{18 b^8} \]
-1/11*a^7*(b*x+a)^11/b^8+7/12*a^6*(b*x+a)^12/b^8-21/13*a^5*(b*x+a)^13/b^8+ 5/2*a^4*(b*x+a)^14/b^8-7/3*a^3*(b*x+a)^15/b^8+21/16*a^2*(b*x+a)^16/b^8-7/1 7*a*(b*x+a)^17/b^8+1/18*(b*x+a)^18/b^8
Time = 0.00 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98 \[ \int x^7 (a+b x)^{10} \, dx=\frac {a^{10} x^8}{8}+\frac {10}{9} a^9 b x^9+\frac {9}{2} a^8 b^2 x^{10}+\frac {120}{11} a^7 b^3 x^{11}+\frac {35}{2} a^6 b^4 x^{12}+\frac {252}{13} a^5 b^5 x^{13}+15 a^4 b^6 x^{14}+8 a^3 b^7 x^{15}+\frac {45}{16} a^2 b^8 x^{16}+\frac {10}{17} a b^9 x^{17}+\frac {b^{10} x^{18}}{18} \]
(a^10*x^8)/8 + (10*a^9*b*x^9)/9 + (9*a^8*b^2*x^10)/2 + (120*a^7*b^3*x^11)/ 11 + (35*a^6*b^4*x^12)/2 + (252*a^5*b^5*x^13)/13 + 15*a^4*b^6*x^14 + 8*a^3 *b^7*x^15 + (45*a^2*b^8*x^16)/16 + (10*a*b^9*x^17)/17 + (b^10*x^18)/18
Time = 0.25 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^7 (a+b x)^{10} \, dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (-\frac {a^7 (a+b x)^{10}}{b^7}+\frac {7 a^6 (a+b x)^{11}}{b^7}-\frac {21 a^5 (a+b x)^{12}}{b^7}+\frac {35 a^4 (a+b x)^{13}}{b^7}-\frac {35 a^3 (a+b x)^{14}}{b^7}+\frac {21 a^2 (a+b x)^{15}}{b^7}+\frac {(a+b x)^{17}}{b^7}-\frac {7 a (a+b x)^{16}}{b^7}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^7 (a+b x)^{11}}{11 b^8}+\frac {7 a^6 (a+b x)^{12}}{12 b^8}-\frac {21 a^5 (a+b x)^{13}}{13 b^8}+\frac {5 a^4 (a+b x)^{14}}{2 b^8}-\frac {7 a^3 (a+b x)^{15}}{3 b^8}+\frac {21 a^2 (a+b x)^{16}}{16 b^8}+\frac {(a+b x)^{18}}{18 b^8}-\frac {7 a (a+b x)^{17}}{17 b^8}\) |
-1/11*(a^7*(a + b*x)^11)/b^8 + (7*a^6*(a + b*x)^12)/(12*b^8) - (21*a^5*(a + b*x)^13)/(13*b^8) + (5*a^4*(a + b*x)^14)/(2*b^8) - (7*a^3*(a + b*x)^15)/ (3*b^8) + (21*a^2*(a + b*x)^16)/(16*b^8) - (7*a*(a + b*x)^17)/(17*b^8) + ( a + b*x)^18/(18*b^8)
3.2.27.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86
method | result | size |
gosper | \(\frac {1}{8} a^{10} x^{8}+\frac {10}{9} a^{9} b \,x^{9}+\frac {9}{2} a^{8} b^{2} x^{10}+\frac {120}{11} a^{7} b^{3} x^{11}+\frac {35}{2} a^{6} b^{4} x^{12}+\frac {252}{13} a^{5} b^{5} x^{13}+15 a^{4} b^{6} x^{14}+8 a^{3} b^{7} x^{15}+\frac {45}{16} a^{2} b^{8} x^{16}+\frac {10}{17} a \,b^{9} x^{17}+\frac {1}{18} b^{10} x^{18}\) | \(113\) |
default | \(\frac {1}{8} a^{10} x^{8}+\frac {10}{9} a^{9} b \,x^{9}+\frac {9}{2} a^{8} b^{2} x^{10}+\frac {120}{11} a^{7} b^{3} x^{11}+\frac {35}{2} a^{6} b^{4} x^{12}+\frac {252}{13} a^{5} b^{5} x^{13}+15 a^{4} b^{6} x^{14}+8 a^{3} b^{7} x^{15}+\frac {45}{16} a^{2} b^{8} x^{16}+\frac {10}{17} a \,b^{9} x^{17}+\frac {1}{18} b^{10} x^{18}\) | \(113\) |
norman | \(\frac {1}{8} a^{10} x^{8}+\frac {10}{9} a^{9} b \,x^{9}+\frac {9}{2} a^{8} b^{2} x^{10}+\frac {120}{11} a^{7} b^{3} x^{11}+\frac {35}{2} a^{6} b^{4} x^{12}+\frac {252}{13} a^{5} b^{5} x^{13}+15 a^{4} b^{6} x^{14}+8 a^{3} b^{7} x^{15}+\frac {45}{16} a^{2} b^{8} x^{16}+\frac {10}{17} a \,b^{9} x^{17}+\frac {1}{18} b^{10} x^{18}\) | \(113\) |
risch | \(\frac {1}{8} a^{10} x^{8}+\frac {10}{9} a^{9} b \,x^{9}+\frac {9}{2} a^{8} b^{2} x^{10}+\frac {120}{11} a^{7} b^{3} x^{11}+\frac {35}{2} a^{6} b^{4} x^{12}+\frac {252}{13} a^{5} b^{5} x^{13}+15 a^{4} b^{6} x^{14}+8 a^{3} b^{7} x^{15}+\frac {45}{16} a^{2} b^{8} x^{16}+\frac {10}{17} a \,b^{9} x^{17}+\frac {1}{18} b^{10} x^{18}\) | \(113\) |
parallelrisch | \(\frac {1}{8} a^{10} x^{8}+\frac {10}{9} a^{9} b \,x^{9}+\frac {9}{2} a^{8} b^{2} x^{10}+\frac {120}{11} a^{7} b^{3} x^{11}+\frac {35}{2} a^{6} b^{4} x^{12}+\frac {252}{13} a^{5} b^{5} x^{13}+15 a^{4} b^{6} x^{14}+8 a^{3} b^{7} x^{15}+\frac {45}{16} a^{2} b^{8} x^{16}+\frac {10}{17} a \,b^{9} x^{17}+\frac {1}{18} b^{10} x^{18}\) | \(113\) |
1/8*a^10*x^8+10/9*a^9*b*x^9+9/2*a^8*b^2*x^10+120/11*a^7*b^3*x^11+35/2*a^6* b^4*x^12+252/13*a^5*b^5*x^13+15*a^4*b^6*x^14+8*a^3*b^7*x^15+45/16*a^2*b^8* x^16+10/17*a*b^9*x^17+1/18*b^10*x^18
Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int x^7 (a+b x)^{10} \, dx=\frac {1}{18} \, b^{10} x^{18} + \frac {10}{17} \, a b^{9} x^{17} + \frac {45}{16} \, a^{2} b^{8} x^{16} + 8 \, a^{3} b^{7} x^{15} + 15 \, a^{4} b^{6} x^{14} + \frac {252}{13} \, a^{5} b^{5} x^{13} + \frac {35}{2} \, a^{6} b^{4} x^{12} + \frac {120}{11} \, a^{7} b^{3} x^{11} + \frac {9}{2} \, a^{8} b^{2} x^{10} + \frac {10}{9} \, a^{9} b x^{9} + \frac {1}{8} \, a^{10} x^{8} \]
1/18*b^10*x^18 + 10/17*a*b^9*x^17 + 45/16*a^2*b^8*x^16 + 8*a^3*b^7*x^15 + 15*a^4*b^6*x^14 + 252/13*a^5*b^5*x^13 + 35/2*a^6*b^4*x^12 + 120/11*a^7*b^3 *x^11 + 9/2*a^8*b^2*x^10 + 10/9*a^9*b*x^9 + 1/8*a^10*x^8
Time = 0.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.99 \[ \int x^7 (a+b x)^{10} \, dx=\frac {a^{10} x^{8}}{8} + \frac {10 a^{9} b x^{9}}{9} + \frac {9 a^{8} b^{2} x^{10}}{2} + \frac {120 a^{7} b^{3} x^{11}}{11} + \frac {35 a^{6} b^{4} x^{12}}{2} + \frac {252 a^{5} b^{5} x^{13}}{13} + 15 a^{4} b^{6} x^{14} + 8 a^{3} b^{7} x^{15} + \frac {45 a^{2} b^{8} x^{16}}{16} + \frac {10 a b^{9} x^{17}}{17} + \frac {b^{10} x^{18}}{18} \]
a**10*x**8/8 + 10*a**9*b*x**9/9 + 9*a**8*b**2*x**10/2 + 120*a**7*b**3*x**1 1/11 + 35*a**6*b**4*x**12/2 + 252*a**5*b**5*x**13/13 + 15*a**4*b**6*x**14 + 8*a**3*b**7*x**15 + 45*a**2*b**8*x**16/16 + 10*a*b**9*x**17/17 + b**10*x **18/18
Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int x^7 (a+b x)^{10} \, dx=\frac {1}{18} \, b^{10} x^{18} + \frac {10}{17} \, a b^{9} x^{17} + \frac {45}{16} \, a^{2} b^{8} x^{16} + 8 \, a^{3} b^{7} x^{15} + 15 \, a^{4} b^{6} x^{14} + \frac {252}{13} \, a^{5} b^{5} x^{13} + \frac {35}{2} \, a^{6} b^{4} x^{12} + \frac {120}{11} \, a^{7} b^{3} x^{11} + \frac {9}{2} \, a^{8} b^{2} x^{10} + \frac {10}{9} \, a^{9} b x^{9} + \frac {1}{8} \, a^{10} x^{8} \]
1/18*b^10*x^18 + 10/17*a*b^9*x^17 + 45/16*a^2*b^8*x^16 + 8*a^3*b^7*x^15 + 15*a^4*b^6*x^14 + 252/13*a^5*b^5*x^13 + 35/2*a^6*b^4*x^12 + 120/11*a^7*b^3 *x^11 + 9/2*a^8*b^2*x^10 + 10/9*a^9*b*x^9 + 1/8*a^10*x^8
Time = 0.31 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int x^7 (a+b x)^{10} \, dx=\frac {1}{18} \, b^{10} x^{18} + \frac {10}{17} \, a b^{9} x^{17} + \frac {45}{16} \, a^{2} b^{8} x^{16} + 8 \, a^{3} b^{7} x^{15} + 15 \, a^{4} b^{6} x^{14} + \frac {252}{13} \, a^{5} b^{5} x^{13} + \frac {35}{2} \, a^{6} b^{4} x^{12} + \frac {120}{11} \, a^{7} b^{3} x^{11} + \frac {9}{2} \, a^{8} b^{2} x^{10} + \frac {10}{9} \, a^{9} b x^{9} + \frac {1}{8} \, a^{10} x^{8} \]
1/18*b^10*x^18 + 10/17*a*b^9*x^17 + 45/16*a^2*b^8*x^16 + 8*a^3*b^7*x^15 + 15*a^4*b^6*x^14 + 252/13*a^5*b^5*x^13 + 35/2*a^6*b^4*x^12 + 120/11*a^7*b^3 *x^11 + 9/2*a^8*b^2*x^10 + 10/9*a^9*b*x^9 + 1/8*a^10*x^8
Time = 0.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int x^7 (a+b x)^{10} \, dx=\frac {a^{10}\,x^8}{8}+\frac {10\,a^9\,b\,x^9}{9}+\frac {9\,a^8\,b^2\,x^{10}}{2}+\frac {120\,a^7\,b^3\,x^{11}}{11}+\frac {35\,a^6\,b^4\,x^{12}}{2}+\frac {252\,a^5\,b^5\,x^{13}}{13}+15\,a^4\,b^6\,x^{14}+8\,a^3\,b^7\,x^{15}+\frac {45\,a^2\,b^8\,x^{16}}{16}+\frac {10\,a\,b^9\,x^{17}}{17}+\frac {b^{10}\,x^{18}}{18} \]